sum_(n=3)^(oo)(Ln n)/(sqrtn) is this divergent or convergent?

Determine Convergence: We need to determine if the series ∑ from n=3 to ∞ of (Ln n)/(√n) is convergent or divergent. We can use the comparison test or integral test to analyze this series. Integral Test: Let's use the integral test. Consider the integral from 3 to ∞ of (Ln x)/(√x) dx. We need to check if this integral converges or diverges. Substitute u: Substitute u = √x, then du = (1/(2√x)) dx, so dx = 2√x du. When x = 3, u = √3, and when x = ∞, u = ∞. The integral becomes ∫ from √3 to ∞ of (Ln u^2)/(u) * 2u du. Simplify Integral: Simplify the integral: ∫ from √3 to ∞ of 2u Ln(u^2)/u du = ∫ from √3 to ∞ of 2 Ln(u^2) du.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

$\sum_{n=3}^{\infty}\frac{\ln n}{\sqrt{n}}$ is this divergent or convergent?

View step-by-step help

Home

Math Problems

Precalculus

Convergent and divergent geometric series

Full solution

\sum_{n=3}^{\infty}\frac{\ln n}{\sqrt{n}}

is this divergent or convergent?

Determine Convergence: We need to determine if the series $\sum_{n=3}^{\infty} \frac{\ln n}{\sqrt{n}}$ is convergent or divergent. We can use the comparison test or integral test to analyze this series.
Integral Test: Let's use the integral test. Consider the integral from $3$ to $\infty$ of $(\ln x)/(\sqrt{x}) \, dx$ . We need to check if this integral converges or diverges.
Substitute u: Substitute $u = \sqrt{x}$ , then $du = \frac{1}{2\sqrt{x}} dx$ , so $dx = 2\sqrt{x} du$ . When $x = 3$ , $u = \sqrt{3}$ , and when $x = \infty$ , $u = \infty$ . The integral becomes $\int_{\sqrt{3}}^{\infty} \frac{\ln u^2}{u} \cdot 2u du$ .
Simplify Integral: Simplify the integral: $\int_{\sqrt{3}}^{\infty} \frac{2u \ln(u^2)}{u} du = \int_{\sqrt{3}}^{\infty} 2 \ln(u^2) du$ .